Ta có:\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{10.10}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}< 1\)

=>A<1

\(\text{Ta có: }\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};.....;\frac{1}{10.10}< \frac{1}{9.10}\)

 \(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\)

\(\Rightarrow A< 1-\frac{1}{10}\)

\(\Rightarrow A< \frac{9}{10}< 1\)

A = 1/2.2 + 1/3.3 + 1/4.4 + ... + 1/10.10

A < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/9.10

A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10

A < 1 - 1/10 < 1

=> A < 1

S= 1/2 - 1/2 + 1/3 - 1/3 + 1/4 - 1/4 +...+ 1/50 - 1/50

S=       0     +       0      +      0      +...+        0

S=  0

\(S=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{49.49}+\frac{1}{50.50}\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{50}-\frac{1}{50}\)

\(=0+0+...+0\)

\(=0\)

Đề bài sai rồi!Riêng 1/(2.2) đã bằng 1/4 rùi thì tổng trên phải lớn hơn 1/4 chứ!

Bạn Phạm Gia Bảo nói đúng đấy

Bạn nên sửa đề bài đi

Thấy :            \(\frac{1}{1.1!}=\frac{1}{1}\)

                       \(\frac{1}{2.2!}=\frac{1}{4}\)

                       \(\frac{1}{3.3!}< \frac{1}{1.2.3}\)( Vì 3.3! > 1.2.3 )

                         ...

                       \(\frac{1}{2019.2019!}< \frac{1}{2017.2018.2019}\)( vì 2019.2019! < 2017.2018.2019)

Cộng từng vế có :

  \(\frac{1}{3.3!}+\frac{1}{4.4!}+...+\frac{1}{2019.2019!}< \frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)

\(\Rightarrow\frac{1}{1.1!}+\frac{1}{2.2!}+...+\frac{1}{2019.2019!}< \frac{1}{1}+\frac{1}{4}+\frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)

\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right):2\)

\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2018.2019}\right):2\)

\(\Rightarrow C< \frac{3}{2}-\frac{1}{2.2018.2019}\)

Vì \(\frac{1}{2.2018.2019}>0\Rightarrow C< \frac{3}{2}\)

Giúp mình bài này với

\(\frac{1}{2.2}+\frac{1}{4.4}+\frac{1}{6.6}+...+\frac{1}{200.200}\)

\(=\frac{1}{4}\left(1+\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)\)

\(< \frac{1}{4}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(=\frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{4}\left(1+1-\frac{1}{100}\right)=\frac{1}{4}\left(2-\frac{1}{100}\right)=\frac{1}{2}-\frac{1}{400}< \frac{1}{2}\)

Cảm ơn bạn

Đặt \(A=\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{100.100}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A< 1-\frac{1}{100}\)

\(\Rightarrow A< \frac{99}{100}\)

Mà \(\frac{99}{100}< 1\Rightarrow A< \frac{99}{100}< 1\)

\(\Rightarrow A< 1\)